digital controls & digital filters lectures 21 & 22 21-22.pdfanalog filter design digital...

Analog Filter Design Digital Filter Design

Digital Controls & Digital FiltersLectures 21 & 22

M.R. Azimi, Professor

Department of Electrical and Computer EngineeringColorado State University

Spring 2015

M.R. Azimi Digital Control & Digital Filters


Review of Analog Filters-Cont.

Types of Analog Filters:

1 Butterworth

2 Chebyshev

3 Elliptic (Cauer)

4 Bessel

Narrower Transition: (1) Elliptic, (2) Chebyshev, (3) Butterworth, (4) Bessel

More linear phase over PB: (1) Bessel, (2) Butterworth, (3) Chebyshev, (4) Elliptic

The first three filters have a common form of magnitude-squared function:

AN (ω2) = 11+F (ω2)

ω = ωωc

: Normalized frequency (ωc: Cutoff frequency )

1. Low-pass Butterworth Filters

Maximally flat passband region

All-pole filter

For sharp cutoff, order must be large



Butterworth Filters

Here F (ω2) = ω2n and n: order

or AN (ω2) = 1

1+(

ωωc

)2n

ω = 0 =⇒ AN (0) = 1ω = ωc =⇒ AN (ω2

c ) = 12

ω large =⇒ AN (ω2) ≈ 0

Changing jω → s in AN (ω2) gives,

AN (−s2) = 1

1+(

sjωc

)2n

Now, to find pole locations of this filter

1 +(

sjωc

)2n

= 0 =⇒ sk = (−1)1/2njωc

But, using (−1)1/N = ej(2k−1)π/N , k ∈ [0, N − 1], we get

sk = ej(2k−1)π/2njωc = sin (2k−1)π2n

+ j cos (2k−1)π2n

jωc, k ∈ [0, 2n− 1]There are n LH poles and n RH poles. The LH poles are the ones chosen according tothe spectral factorization (or use tables).



Butterworth Filters-Cont.

Disadvantages:

1 Wide transition region (for low order filters).

2 Accuracy of approximation cannot be uniformly distributed in PB and SB regions.

Example:Design a Butterworth LPF, HA(s), that has attenuation of at least 10dB at ω = 2ωcwhere ωc = 2.5× 103

AN (ω2) = 11+ω2n

AN (22) = 11+22n

−10 log10 |AN (22)| ≥ 10 =⇒ (1 + 22n) ≥ 10 =⇒ n = 1.58 =⇒ n = 2

From table: HN (s) = 1s2+1.414s+1

Denormalize, s→ s/ωc

HA(s) = HN ( sωc

) = 11.6×10−7s2+5.657×10−4s+1



Butterworth Filters-Cont.

Short-Cut Method:

The required filter order for providing a magnitude of 1/A (not in dB) at frequency ωcan be obtained using,

n = log10(A2−1)

2 log10( ωωc

)

For previous example,

−20 log10 1/A = 10 dB =⇒ A = 3.3

n = log10(3.32−1)

2 log10 2= 1.653 =⇒ n = 2




2. Lowpass Chebyshev Filters

Sharper cutoff or narrower transition region.

Magnitude of error is equiripple over the passband (Type I) or over the stopband(Type II).

Type I Chebyshev: Magnitude-squared function is given byAN (ω2) = |HN (jω)|2 = 1

1+F (ω2)

where F (ω2) = ε2C2n(ω)

ε: Ripple factor (determines ripple in passband)Cn(ω): nth order Chebyshev polynomialCn(ω) = cos (n cos−1 ω) for 0 ≤ ω ≤ 1 (passband)Cn(ω) = cosh (n cosh−1 ω) for ω > 1 (stopband)For any n, 0 ≤ |CN (ω)| ≤ 1 for 0 ≤ ω ≤ 1

|CN (ω)| > 1 for ω > 1n = 1 =⇒ C1(ω) = ωn = 2 =⇒ C2(ω) = 2ω2 − 1n = 3 =⇒ C3(ω) = 4ω3 − 3ω...i.e. Cn(ω) is an odd/even polynomial when n is odd/even.



Chebyshev Filters-Cont.

Chebyshev polynomials can be generated recursively using,

Cn+1(ω) + Cn−1(ω) = 2ωCn(ω)

For n : even, C2n(0) = 1 =⇒ A(0) = 1

1+ε2

For n : odd, C2n(0) = 0 =⇒ A(0) = 1

max |HN (jω)| = Amax = 1: Max gain in passbandmin |HN (jω)| = Amin = 1√

1+ε2: Min gain in passband

Define:rdB = 20 log10

AmaxAmin

= 10 log10 (1 + ε2) : Loss in passband

r(peak-to-peak) = Amax −Amin = 1− 1√1+ε2




To find pole locations of the Chebyshev filter, change ω = s/j and set thedenominator to zero,

1 + ε2C2n(s/j) = 0 =⇒ cos (n cos−1( s

j)) = ± j

ε

But, using sk = σk + jωk, we get

σk = − sinhϕ sin[ (2k−1)π2n

] k ∈ [0, 2n− 1]

ωk = coshϕ cos[ (2k−1)π2n

] k ∈ [0, 2n− 1]

wheresinhϕ = γ−γ−1

2, coshϕ = γ+γ−1

2, and

γ = (1+√

1+ε2

ε)1/n.

It is interesting to note that from equations for σk and ωk, we have,

σ2k

sinh2 ϕ+

ω2k

cosh2 ϕ= 1

i.e. the poles are located on an ellipse in the s-plane.




Example:

Design a Chebyshev Type I filter with 1dB ripple in passband and attenuation of atleast 20dB at ω = 2ωc, fc = 3kHz.

r = 10 log10 (1 + ε2) = 1 =⇒ ε = .05088

−10 log101

1+ε2C2n(2)≥ 20

Since in stopband, ε2C2n(2) 1

Thus,10 log10 ε

2 + 10 log10 C2n(2) ≥ 20

log10 [cosh(n cosh−1 2)] ≥ 1− log10 ε = 1.2935 =⇒ n cosh−1 2 ≥ 3.67

or n ≥ 2.8 =⇒ n = 3

Use Table 11.4 (for r = 1dB)

HN (s) = 0.4913s3+0.9883s2+1.2384s+0.4913

Denormalize: ωc = 2π × 3× 103 = 6π × 103Rad/Sec

s→ sωc

HA(s) = 0.49131.493×10−13s3+2.78×10−9s2+6.569×10−5s+0.4913




Short-Cut Method:If the magnitude of minimum stopband loss is 1/A at frequency ωr, then the order can

be found using n =log10(g+

√g2−1)

log10(ωr+√ω2r−1)

where g =√

A2−1ε2

3. Elliptic Filters:Characteristics are:

1 Equiripple both in passband and stopband.

2 Excellent transition region.

The magnitude-squared function for elliptic filters is:

A(ω2) = 11+ε2R2

n(ω,κ1)

Rn(ω, κ1) : Chebyshev rational function with roots that are related to Jacobi ellipticfunctions hence called elliptic filters.κ1 , ε/

√A2 − 1: selectivity factor



Elliptic Filters

Define transition ratio p = ωcωr

. Then the order of the elliptic filter to meet ε, A (asdefined before) and ωc, ωr specs is:

n =K(p)K(

√1−κ2

1)

K(κ1)K(√

1−p2)

K(.): Complete elliptic integral of the 1st kind,

K(x) =1

0

1

(1−t2)1/2(1−xt2)1/2dt

Remarks:

1 In this filter, ε (ripple factor) determines the passband ripple, while thecombination of ε and κ1 specify the stopband ripple.

2 To obtain equiripple in both passband and stopband elliptic filters must havepoles on jω axis.

3 When κ1 →∞ the elliptic rational function becomes a Chebyshev polynomial,and therefore the filter becomes a Chebyshev Type I filter, with ripple factor ε.

Disadvantages:

1 Nonlinear phase characteristics.

2 Delay varies with frequency.




4. Bessel Filters:

Excellent phase characteristic (nearly linear).

Wide transition region (disadvantage).

Cutoff Frequency changes as function of n (disadvantage).

All-pole filter with transfer function,HA(s) = d0

Bn(s)

where Bn(s): nth order Bessel Polynomial

Bn(s) =n∑i=0

disi

di = (2n−i)!2n−ii!(n−i)! , i ∈ [0, n]

Bessel polynomials satisfy, Bn(s) = (2n− 1)Bn−1(s) + s2Bn−2(s)

with B0(s) = 0, B1(s) = s+ 1

Cutoff Frequency varies as function of order n, ωc ≈ d1/n0



Recursive Digital Filter Design Using Bilinear z-Mapping

Idea: Analog systems are implemented using integrators, multipliers, and adders. Mapevery integrator HI(s) = 1/s (or hI(t) = us(t)) to the digital domain.

Convolution integral: y(t) =t

0

x(τ)hI(t− τ)dτ

Let 0 < t1 < t2, then

y(t2) =t2

0

x(τ)hI(t− τ)dτ

y(t1) =t1

0

x(τ)hI(t− τ)dτ

y(t2)− y(t1) =t2

t1

x(τ)dτ =Area

≈ (t2−t1)2

[x(t1) + x(t2)] Trapezoidal rule of integration.

Let t1 = (n− 1)T, t2 = nT

y(nT )− y((n− 1)T ) = T2

[x((n− 1)T ) + x(nT )]



Bilinear z-Mapping-Cont.

Take z-transform:

Y (z)− z−1Y (z) = T2

[z−1X(z) +X(z)]

HI(z) = Y (z)X(z)

= T2

(1+z−1)

1−z−1 = T2z+1z−1

Thus the mapping from s-domain to z-domain is:

1/s→ T2

(z+1)(z−1)

s→ 2T

(1−z−1)

(1+z−1): Bilinear z-mapping

Once HA(s) prototype is described:

HD(z) = HA(s)|s= 2

T(1−z−1)

(1+z−1)

Properties of Bilinear z-Mapping:

Recall: s = 2T

(z−1)z+1)

or z = 2/T+s2/T−s

Also z = rejΩ, and s = σ + jω which gives



Bilinear z-Mapping-Properties

rejΩ = (2/T+σ)+jω(2/T−σ)−jω

Then equations for r and Ω become,

r =[

(2/T+σ)2+ω2

(2/T−σ)2+ω2

]1/2

Ω = tan−1 ω2T +σ

+ tan−1 ω2T −σ

Right half of s-plane: σ > 0 =⇒ r > 1: outside unit circle

On jω axis: σ = 0 =⇒ r = 1: on unit circle

Left half of s-plane: σ < 0 =⇒ r < 1: inside the unit circle

=⇒ Stable HA(s) is mapped to stable HD(z).

Additionally, it is obvious that HD(z) will have real coefficients. However, thereis an issue with this mapping called warping.



Bilinear z-Mapping-Warping Effects

For simplicity, let σ = 0 in equation of Ω,

Ω = 2 tan−1 ω2/T = 2 tan−1 ωT

2

or ω = 2T tan Ω

2 : Nonlinear relation between Ω, ω

Let Ω = ΩT (Normalized Ω)

ω = 2T tan ΩT

2

When ΩT2 < 0.15, tan ΩT

2 ≈ΩT2 =⇒ ω ≈ Ω

i.e. approximately linear portion of tangent curve.

Thus, If all essential frequencies (ωc, ωr) are below Ω = 0.3T , no prewarping is

needed since minimal warping effects and ω ≈ Ω.

If not, prewarp the analog filter before bilinear z-mapping.



Bilinear z-Mapping-Warping Effects

Note: Warping affects both magnitude and phase responses. Although, the amplitudedistortion caused by warping can be remedied by prewarping, little can be done tolinearize the phase except maybe using equalization methods.



Bilinear z-Mapping-Prewarping

Procedure:If all essential frequencies are above Ω = 0.3

T, then

1 Map all essential digital frequencies Ωi to corresponding analog frequencies using

ωi = 2T

tan ΩiT2

2 Design HA(s) using these analog frequencies ωi’s.

3 Map HA(s) to HD(z) using Bilinear z-mapping.

Example:

Design a digital Butterworth filter that meets:

1 T = 50 µ sec.

2 fc = 4 kHz or Ωc = 25.13× 103 Rad/sec

3 At 2Ωc attenuation is ≥ 15dB

Since Ωc 0.3T≈ 6000 =⇒ Prewaring is needed

We prewarp both Ωc and 2Ωc, using



Bilinear z-Mapping-Cont.

ωc = 2T

tan ΩcT2≈ 29.1× 103 Rad/sec

ω15dB = 2T

tan 2ΩcT2

= 123.1× 103 Rad/sec

We now use these frequencies to design HA(s)

|HA(jω)|2 = 1

1+(

ωωc

)2n

−10 log10

∣∣∣∣ 1

1+( 123.129.1 )2n

∣∣∣∣ ≥ 15dB =⇒ n ≥ 1.2 =⇒ n = 2

From Table 11.2:

HN (s) = 1s2+1.4142s+1

Denormalize s→ s/ωc

HA(s) = (29×103)2

s2+41.15×103s+(29×103)2

HD(z) = HA(s)|s= 2

T1−z−1

1+z−1

= 2.117+4.234z−1+2.117z−2

10.232−3.766z−1+2.002z−2


digital controls & digital filters lectures 21 & 22 21-22.pdfanalog filter design digital...

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